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[Гост]

[tex]\overline{AB}[/tex] : [tex]\overline{BC}[/tex] : [tex]\overline{CD}[/tex] : [tex]\overline{DA}[/tex] = 2 : 7 : 3 : 6
AB [tex]\cap[/tex] CD = M
AC [tex]\cap[/tex] BD = N
През точка C е построена допирателна CP към окръжността.
Намерете големина на [tex]\angle[/tex]BMC, [tex]\angle[/tex]ANB и [tex]\angle[/tex]ACP.

[Гост]

Ето чертеж:

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$\stackrel{\Large\frown}{AB}=2x,\stackrel{\Large\frown}{BC}=7x,\stackrel{\Large\frown}{CD}=3x,\stackrel{\Large\frown}{DA}=6x$
$\stackrel{\Large\frown}{AB}+\stackrel{\Large\frown}{BC}+\stackrel{\Large\frown}{CD}+\stackrel{\Large\frown}{DA}=360^\circ$
$(2+7+3+6)x=360^\circ$
$18x=360^\circ$
$x=20^\circ$

$\Rightarrow\stackrel{\Large\frown}{AB}=40^\circ,\stackrel{\Large\frown}{BC}=140^\circ,\stackrel{\Large\frown}{CD}=60^\circ,\stackrel{\Large\frown}{DA}=120^\circ$

$\Rightarrow\stackrel{\Large\frown}{AB}+\Rightarrow\stackrel{\Large\frown}{BC}=180^\circ\Rightarrow AC$ - диаметър.

$\angle BMC=\frac{1}{2}\left(\stackrel{\Large\frown}{BC}-\stackrel{\Large\frown}{AD}\right)=\frac{140^\circ-120^\circ}{2}=10^\circ$

$\angle ANB=\frac{1}{2}\left(\stackrel{\Large\frown}{AB}+\stackrel{\Large\frown}{CD}\right)=\frac{40^\circ+60^\circ}{2}=50^\circ$

$\angle ACP=90^\circ$ (Поради това, че АС е диаметър, а СР - допирателна.)